A proof that $1=2$. May I know why it’s false?

Question

im not good at formatting. May I know why it’s false

$x^2= x+x+x+\ldots(x\; \text{times})$

apply derivative on both sides

$=> \frac{d}{dx}(x^2)=\frac{d}{dx}(x+x+x+\ldots(x\; \text{times}))$

$=> 2x=1+1+1+\ldots(x\; \text{times})$

$=> 2x=x$

$=> 2=1$

How do you make sense of $x+x+x+..(x\text{ times})$ when $x$ is not a whole number? — 2'5 9'2, Apr 02 '13 at 04:59
But the real problem is that your right-hand derivative has not considered the "$x$ times". The right-hand side depends on that; so the derivative should do something with it. — 2'5 9'2, Apr 02 '13 at 05:00
Also http://math.stackexchange.com/questions/1096/where-is-the-flaw-in-this-argument-of-a-proof-that-1-2-derivative-of-repeated — MJD, Apr 02 '13 at 06:37
You can't differentiate a function that isn't continuous since it isn't define when $x$ isn't a whole number, remember, non-continuous means non-differentiable. — Hakim, May 12 '14 at 21:03

score 2 · Accepted Answer · answered Apr 02 '13 at 05:02

You cannot add '$x$', $x$ number of times without knowing what $x$ exactly is. Hence the differentiation on the RHS has no real meaning.

This fails for many cases like:

$1)$ '$x$' is not an integer.

$2)$ Dependency of $x$ on $x$ which you are not considering when differentiating on RHS.

score 0 · Answer 2 · answered Apr 02 '13 at 05:00

0

This has been asked twice or thrice already. Two problems might be pointed out

$(1)$ "$x$ times" is ill-defined for non integer values of $x$; and even so

$(2)$ "$x$ times" is also a function of $x$, which you're ignoring

answered Apr 02 '13 at 05:00

Pedro

2 Answers2